已知函数fx=2√3sinXcosX+2cos²X+m在区间[0,π/2]上最大值为2求常数m值 2,在三角形ABC中角ABC所对的边长abc若F(A)=1,sinB=3sinC ,S三角形ABC=3√3/4求边长a
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![已知函数fx=2√3sinXcosX+2cos²X+m在区间[0,π/2]上最大值为2求常数m值 2,在三角形ABC中角ABC所对的边长abc若F(A)=1,sinB=3sinC ,S三角形ABC=3√3/4求边长a](/uploads/image/z/1615636-28-6.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0fx%3D2%E2%88%9A3sinXcosX%EF%BC%8B2cos%26%23178%3BX%2Bm%E5%9C%A8%E5%8C%BA%E9%97%B4%5B0%2C%CF%80%2F2%EF%BC%BD%E4%B8%8A%E6%9C%80%E5%A4%A7%E5%80%BC%E4%B8%BA2%E6%B1%82%E5%B8%B8%E6%95%B0m%E5%80%BC+2%2C%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%E8%A7%92ABC%E6%89%80%E5%AF%B9%E7%9A%84%E8%BE%B9%E9%95%BFabc%E8%8B%A5F%EF%BC%88A%EF%BC%89%3D1%2CsinB%3D3sinC+%2CS%E4%B8%89%E8%A7%92%E5%BD%A2ABC%3D3%E2%88%9A3%EF%BC%8F4%E6%B1%82%E8%BE%B9%E9%95%BFa)
已知函数fx=2√3sinXcosX+2cos²X+m在区间[0,π/2]上最大值为2求常数m值 2,在三角形ABC中角ABC所对的边长abc若F(A)=1,sinB=3sinC ,S三角形ABC=3√3/4求边长a
已知函数fx=2√3sinXcosX+2cos²X+m在区间[0,π/2]上最大值为2求常数m值 2,在三角形ABC中角ABC所对的边长abc若F(A)=1,sinB=3sinC ,S三角形ABC=3√3/4求边长a
已知函数fx=2√3sinXcosX+2cos²X+m在区间[0,π/2]上最大值为2求常数m值 2,在三角形ABC中角ABC所对的边长abc若F(A)=1,sinB=3sinC ,S三角形ABC=3√3/4求边长a
F(x)=2√3sinXcosX+2cos²X+m=√3sin2x+cos2x-1+m=2sin(2x+π/6)-1+m
0≤x≤π/2, 0≤2x≤π, π/6≤2x+π/6≤7π/6 最大值=2-1+m=2, 所以m=1
F(A)=1 , 2sin(2A+π/6)=1, sin(2A+π/6)=1/2, 2A+π/6=2kπ+π/6或2kπ+5π...
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F(x)=2√3sinXcosX+2cos²X+m=√3sin2x+cos2x-1+m=2sin(2x+π/6)-1+m
0≤x≤π/2, 0≤2x≤π, π/6≤2x+π/6≤7π/6 最大值=2-1+m=2, 所以m=1
F(A)=1 , 2sin(2A+π/6)=1, sin(2A+π/6)=1/2, 2A+π/6=2kπ+π/6或2kπ+5π/6,所以A=π/3
已知sinB=3sinC 所以b=3c
1/2bcsinA=1/2*3c^2*√3/2=3√3/4, 所以c=1, b=3, a^2=1+9-2*1*3*√3/2=10-3√3, a=√[10-3√3]
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